# An interesting phenomenon of $C^*$-tensor product

On the algebraic tensor product space of $C^*$-algebra, I try to find an example whose maximal $C^*$-norm is not the minimal $C^*$-norm, but it seems as it is impossible to do this because the finite $C^*$-algebra is nuclear and so does the AF-algebra, so the $C^*$-norm has an interesting phenomenon, the maxiamal $C^*$-norm must be equal to minimal $C^*$-norm except one $C^*$-norm has no definition (the $C^*$-tensor product space is smaller).

Am I right? Does any name of interesting phenomenon and any this books or papers discuss this phenomenon?

• I cannot really understand what you are saying. Which algebra is finite? And what do you mean by saying that a C*-algebra is "finite"? Which norm has no definition? – Martin Argerami May 17 '14 at 20:17
• Finite means finite dimension., given an element of algebraic tensor product, it is only dependent on finite elements, oh, maybe I make a mistake here, finite element can generate an infinite C*-algebra. So maybe I should change this question, to find an example whose maximal C∗-norm is not the minimal C∗-norm. – Strongart May 18 '14 at 6:15
• in tha algebraic tensor product, the elemant is like Σa$_i$⊙b$_i$, here is finite sum, we can choose the C*-Algebra generated by finite a$_i$. If a C^algebra generated by finite element must be AF-algebra（so it is nuclear), maybe we can save this conclusion. – Strongart May 18 '14 at 13:43

if $A$ is a finite-dimensional and $B$ is any C$^*$-algebra, then the algebraic tensor $A\odot B$ is already a C$^*$-algebra, and so there is a single possible tensor norm.
On the other hand, singly generated is not enough to make a C$^*$-algebra AF, nor nuclear; not even exact (see the answer to this question). For any such $A$, as it is non-nuclear, there exists $B$ with $A\otimes_\min B\ne A\otimes_\max B$.
• Thanks, maybe find an exact x such that ‖x‖$_max$=‖x‖$_min$ is not simple, the nuclear C*-algebra is large class. – Strongart May 19 '14 at 14:22